Question 391285
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The quadratic formula gives the general solution of a quadratic equation.  The formula was developed by completeing the square on the general quadratic equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ax^2\ +\ bx\ +\ c\ =\ 0]


Move the constant term to the RHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ax^2\ +\ bx\ =\ -\ c]


Divide by the lead coefficient:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ \frac{b}{a}x\ =\ -\ \frac{c}{a}]


Divide the coefficient on the 1st degree term by 2, square the result, then add that result to both sides of the equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ \frac{b}{a}x\ +\ \frac{b^2}{4a^2}\ =\  \frac{b^2}{4a^2}\ -\ \frac{c}{a}]


Apply the LCD and complete the sum in the RHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ \frac{b}{a}x\ +\ \frac{b^2}{4a^2}\ =\  \frac{b^2\ -\ 4ac}{4a^2}]


Factor the perfect square in the LHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\ +\ \frac{b}{2a}\right)^2\ =\  \frac{b^2\ -\ 4ac}{4a^2}]


Take the square root remembering to consider both the positive and negative root:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ \frac{b}{2a}\ =\  \pm\sqrt{\frac{b^2\ -\ 4ac}{4a^2}}\ =\ \frac{\pm\sqrt{b^2\ -\ 4ac}}{2a}]


Add *[tex \Large -\frac{b}{2a}] to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-b\ \pm\sqrt{b^2\ -\ 4ac}}{2a}]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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